ĐKXĐ: \(x\ge-2;y\ge0\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x+2}=a\ge0\\\sqrt{y}=b\ge0\end{matrix}\right.\) pt đầu trở thành:

\(a\left(a^2+1\right)=b\left(ab+1\right)\)

\(\Leftrightarrow a^3+a=ab^2+b\)

\(\Leftrightarrow a^3-ab^2+a-b=0\)

\(\Leftrightarrow a\left(a^2-b^2\right)+a-b=0\)

\(\Leftrightarrow a\left(a-b\right)\left(a+b\right)+a-b=0\)

\(\Leftrightarrow\left(a-b\right)\left(a^2+ab+1\right)=0\)

\(\Leftrightarrow a-b=0\) (do \(a^2+ab+1>0;\forall a\ge0;b\ge0\))

\(\Leftrightarrow\sqrt{x+2}=\sqrt{y}\)

\(\Rightarrow y=x+2\)

Thế vào pt dưới:

\(x^2+\left(x+3\right)\left(x+3\right)=x+16\)

\(\Leftrightarrow2x^2+5x-7=0\Rightarrow\left[{}\begin{matrix}x=1\Rightarrow y=3\\x=-\dfrac{7}{2}< -2\left(loại\right)\end{matrix}\right.\)

Từ \(x+y=4\Rightarrow y=4-x\)

Thế vào \(x^2+y^2=10\)

\(\Rightarrow x^2+\left(4-x\right)^2=10\)

\(\Leftrightarrow2x^2-8x+6=0\Rightarrow\left[{}\begin{matrix}x=1\Rightarrow y=3\\x=3\Rightarrow y=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)^2-2xy=65\\xy-\left(x+y\right)=17\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}x+y=u\\xy=v\end{matrix}\right.\) với \(u^2\ge4v\)

\(\Rightarrow\left\{{}\begin{matrix}u^2-2v=65\\v-u=17\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}u^2-2v=65\\v=u+17\end{matrix}\right.\)

Thế pt dưới vào pt trên:

\(\Rightarrow u^2-2\left(u+17\right)=65\)

\(\Leftrightarrow u^2-2u-99=0\Rightarrow\left[{}\begin{matrix}u=11\Rightarrow v=28\\u=-9\Rightarrow v=8\end{matrix}\right.\)

- TH1: \(\left\{{}\begin{matrix}x+y=11\\xy=28\end{matrix}\right.\) \(\Rightarrow\left(x;y\right)=\left(7;4\right);\left(4;7\right)\)

TH2: \(\left\{{}\begin{matrix}x+y=-9\\xy=8\end{matrix}\right.\) \(\Rightarrow\left(x;y\right)=\left(-1;-8\right);\left(-8;-1\right)\)

ĐKXĐ: \(x\ge-1;y\ne\dfrac{1}{2}\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x+1}=u\\\dfrac{1}{2y-1}=v\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}3u+2v=4\\4u-v=9\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3u+2v=4\\8u-2v=18\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}11u=22\\v=4u-9\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}u=2\\v=-1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\sqrt{x+1}=2\\\dfrac{1}{2y-1}=-1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x+1=4\\2y-1=-1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=3\\y=0\end{matrix}\right.\)

a: Xét ΔAOC có AO=AC và \(\widehat{OAC}=90^0\)

nên ΔAOC vuông cân tại A

=>\(\widehat{AOB}=\widehat{AOC}=45^0\)

Xét (O) có

\(\widehat{AOB}\) là góc ở tâm chắn cung AB nhỏ

=>\(sđ\stackrel\frown{AB}\left(lớn\right)=360^0-\widehat{AOB}=315^0\)

b: Xét (O) có

ΔABD nội tiếp

AD là đường kính

Do đó: ΔABD vuông tại B

Xét ΔOAB có \(cosAOB=\dfrac{OA^2+OB^2-AB^2}{2\cdot OA\cdot OB}\)

=>\(\dfrac{R^2+R^2-AB^2}{2\cdot R\cdot R}=cos45=\dfrac{\sqrt{2}}{2}\)

=>\(2R^2-AB^2=R^2\cdot\sqrt{2}\)

=>\(AB^2=R^2\left(2-\sqrt{2}\right)\)

=>\(AB=R\sqrt{2-\sqrt{2}}\)

ΔABD vuông tại B

=>\(BA^2+BD^2=AD^2\)

=>\(BD^2=AD^2-AB^2=4R^2-R^2\left(2-\sqrt{2}\right)=R^2\left(2+\sqrt{2}\right)\)

=>\(BD=R\sqrt{2+\sqrt{2}}\)

Xét ΔABD vuông tại B có \(cosBAD=\dfrac{AB}{AD}=\dfrac{R\sqrt{2-\sqrt{2}}}{2R}=\dfrac{\sqrt{2-\sqrt{2}}}{2}\)

=>\(\widehat{BAD}=67,5^0\)

ΔABD vuông tại B

=>\(\widehat{BAD}+\widehat{BDA}=90^0\)

=>\(\widehat{BDA}=90^0-67,5^0=22,5^0\)

ĐKXĐ: \(x\ne2y;x\ne\dfrac{y}{2}\)

\(\left\{{}\begin{matrix}\dfrac{2}{2x-y}+\dfrac{3}{x-2y}=\dfrac{1}{2}\\\dfrac{2}{2x-y}-\dfrac{1}{x-2y}=\dfrac{1}{18}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{4}{x-2y}=\dfrac{4}{9}\\\dfrac{2}{2x-y}-\dfrac{1}{x-2y}=\dfrac{1}{18}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{x-2y}=\dfrac{1}{9}\\\dfrac{1}{2x-y}=\dfrac{1}{2\left(x-2y\right)}+\dfrac{1}{36}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{x-2y}=\dfrac{1}{9}\\\dfrac{1}{2x-y}=\dfrac{1}{12}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x-2y=9\\2x-y=12\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=5\\y=-2\end{matrix}\right.\)

Từ pt đầu \(\Rightarrow y=2x-1\)

Thế xuống dưới:

\(x^2+x\left(2x-1\right)+2\left(2x-1\right)^2=4\)

\(\Leftrightarrow11x^2-9x-2=0\)

\(\Rightarrow\left[{}\begin{matrix}x=1\Rightarrow y=1\\x=-\dfrac{2}{11}\Rightarrow y=-\dfrac{15}{11}\end{matrix}\right.\)

\(\left\{{}\begin{matrix}x+y+xy=5\\x^2+y^2=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+y+xy=5\\\left(x+y\right)^2-2xy=5\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}x+y=u\\xy=v\end{matrix}\right.\)  với \(u^2\ge4v\)

\(\Rightarrow\left\{{}\begin{matrix}u+v=5\\u^2-2v=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}v=5-u\\u^2-2v=5\end{matrix}\right.\) \(\Rightarrow u^2-2\left(5-u\right)=5\)

\(\Leftrightarrow u^2+2u-15=0\)

\(\Rightarrow\left[{}\begin{matrix}u=3\Rightarrow v=2\\u=-5\Rightarrow v=10\left(loại\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x+y=3\\xy=2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}y=3-x\\xy=2\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}y=3-x\\x\left(3-x\right)=2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}y=3-x\\x^2-3x+2=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=1;y=2\\x=2;y=1\end{matrix}\right.\)

Em kiểm tra lại đề, em chắc là \(y^5\) chứ?

Lời giải:
Theo định lý Bê-du về phép chia đa thức, số dư của  $P(x)$ khi chia $2x-5$ là $P(\frac{5}{2})=\frac{5}{4}(\frac{5}{2})^3+\frac{5}{6}(\frac{5}{2})^2-\frac{21}{4}.\frac{5}{2}+\frac{1}{6}=\frac{377}{32}$